Combinations and the sequencing of DNA 
There are four different nucleotides in DNA, from a mathematical point
  of view, DNA sequences can be represented by words over the four
  letters alphabet $A = \{A, C, G, T\}$. 
(a) If you could write one sequences a second, how long will it take
  you to write all DNA sequences of size 12? Give your answer in the
  format (hours, minutes, seconds).
(b) Of all the DNA sequences of size 12, in how many of them do we see
  both the sequences $ACGTATA$ and $TATAC$? We say that $ACGTATA$ and
  $TATAC$ are subsequences.

My solution for part a: 
$hours = \frac{\frac{12!}{60}}{60} $
My solution for part b:
Put side by side,
$ACGTATA$ and $TATAC$  
could be solved in two ways; 
$ACGTATA$$TATAC$ and $TATAC$$ACGTATA$  
when $TATACGTATA$ is the sequence there are 2 free spaces for DNA nucleotides. So this may be done in $(4*4)(4*4)(4*4)(4*4) = 4096$ ways.
when $ACGTATAC$ is the sequence there are 4 free spaces for DNA nucleotides. So this may be done in $(4*4*4)(4*4*4)(4*4*4)(4*4*4)(4*4*4) = 16777216$ ways.
Please let my know whether or not my solution to the problem is correct. Please correct any mistakes I have made. Thanks in advance for your help!
 A: (a) You can choose one of four letters at 12 positions, which gives $4^{12}$ possible sequences of length 12.
At one sequence per second, it takes $(4660 \mbox{h}, 20\mbox{m}, 16 \mbox{s})$
(b) We have these cases:


*

*$ACGTATA \cdot TATAC$, contributing $1$ word

*$TATAC \cdot ACGTATA$, contributing $1$ word

*$u \cdot ACGTATA \cdot TAC \cdot v$, with $\lvert uv \rvert = 2$, contributing $3\times 4^2 = 48$ words

*$u \cdot ACGTATA \cdot C \cdot v$, with $\lvert uv \rvert = 4$, contributing $5 \times 4^4 = 1280$ words

*$u \cdot TAT \cdot ACGTATA \cdot v$, with $\lvert uv \rvert = 2$, contributing $3\times 4^2 = 48$ words


Where $\cdot$ is the concatenation operator (which like the multiplication can be omitted) and $u, v \in \{ A, C, G, T \}^*$ (finite DNA sequences, including the empty sequence).
The factors result from the choices to split a word $w$ into two substrings $w = u \cdot v$.
This gives $1+1+48+1280+48= 1378$ words.
However my friend Ruby thinks these are only $1370$ words, so we need to eliminate some multiple countings.
Asked about which words are counted more than once and why, she told me:
517: ATATACGTATAC [4] [5]
788: CTATACGTATAC [4] [5]
1059: GTATACGTATAC [4] [5]
1202: TATACGTATACA [4] [5]
1203: TATACGTATACC [4] [5]
1204: TATACGTATACG [4] [5]
1205: TATACGTATACT [4] [5]
1343: TTATACGTATAC [4] [5]

These are the sequences $u \cdot TAT \cdot ACGTATA \cdot C \cdot v$ with $\lvert uv \rvert = 1$, contributing $2 \times 4^1 = 8$ words. This is the intersection of the fourth and fifth case above. 
